statistics – How can two seemingly identical conditional expectations have different values?

Background

Suppose that we are using a simplified spherical model of the Earth’s surface with latitude $$u in (-frac {pi} 2, frac {pi} 2)$$ and longitude $$v in (-pi, pi)$$; then (if the radius is taken to be $$1$$), the surface area element is given by $$mathrm{d}A = cos u mathrm{d}u mathrm{d}v$$. Restricting attention to the hemisphere, $$H$$, where $$u, v in (-frac {pi} 2, frac {pi} 2)$$, a simple map projection from $$H$$ can be obtained by just taking the $$x$$ and $$y$$ coordinates via $$x = cos u sin v$$ and $$y = sin u$$, which is a smooth one-to-one transformation on $$H$$. Now, picking a point with coordinates $$(U, V)$$ on $$H$$ uniformly according to surface area, the joint density of $$U$$ and $$V$$ is $$f_{U, V}(u, v) = frac 1 {2pi} cos u, quad lvert u rvert, lvert v rvert < frac {pi} 2.$$

Question

$$(a)quad$$ Find $$mathbb{E}(lvert sin U rvert mid V = 0)$$.

$$(b)quad$$ Find $$mathbb{E}(lvert Y rvert mid X = 0)$$.

$$(c)quad$$ Observe that $$lvert Y rvert = lvert sin U rvert$$ and the event $${X = 0}$$ is exactly the same as the event $${V = 0}$$. How is it possible that $$mathbb{E}(lvert Y rvert mid X = 0) neq mathbb{E}(lvert sin U rvert mid V = 0)$$?

My working

I have omitted intermediate steps and only shown the essential parts.

$$(a)$$

begin{aligned} because f_{U mid V = v}(u) & = frac 1 2 cos u,quad lvert u rvert, lvert v rvert < frac pi 2 \(5 mm) therefore mathbb{E}(lvert sin U rvert mid V = 0) & = int^{infty}_{-infty} lvert sin u rvert left(frac 1 2 cos uright) mathrm{d}u \(5 mm) & = int^{frac pi 2}_0 sin u cos u mathrm{d}u \(5 mm) & = frac 1 2 end{aligned}

$$(b)$$

begin{aligned} \(5 mm) because f_{X, Y}(x, y) & = frac 1 {2 pi sqrt{1 – y^2 – x^2}}, quad x^2 + y^2 < 1 \(5 mm) therefore f_{Y mid X = x}(y) & = frac {frac 1 {2 pi sqrt{1 – y^2 – x^2}}} {int^{sqrt{1 – x^2}}_{-sqrt{1 – x^2}} frac 1 {2 pi sqrt{1 – y^2 – x^2}} mathrm{d}y} \(5 mm) & = frac 1 {pi sqrt{1 – y^2 – x^2}}, quad x^2 + y^2 < 1 \(5 mm) implies mathbb{E}(lvert Y rvert mid X = 0) & = int^{infty}_{-infty} frac {lvert y rvert} {pi sqrt{1 – y^2}} mathrm{d}y \(5 mm) & = frac 2 pi int^1_0 frac y {pi sqrt{1 – y^2}} mathrm{d}y \(5 mm) & = frac 2 pi end{aligned}

I think my answers to $$(a)$$ and $$(b)$$ are correct, but I am not sure how to approach $$(c)$$. Any intuitive explanations will be greatly appreciated! Also, if my answers to $$(a)$$ and/or $$(b)$$ are wrong, please do point out my mistakes 🙂